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MATH 3150
Homework Problem Set 1 Solutions
(1) (#6) Let a, b, c, d be rational numbers and x an irrational number such that cx + d 6= 0.
Prove that (ax + b)/(cx + d) is irrational if and only if ad 6= bc.
Proof. We can prove equivalently that (ax + b)/(cx + d) is rational if and only if ad = bc.
First, we show ad = bc ⇒ (ax + b)/(cx + d) ∈ Q.
• If a = 0 then bc = 0. If b = 0, we have (ax + b)/(cx + d) = 0 is rational; if c = 0, since
cx + d 6= 0, d 6= 0 and (ax + b)/(cx + d) = b/d ∈ Q since b, d ∈ Q and Q is a field.
• If a 6= 0, note that c 6= 0 (otherwise, d = 0 and cx + d = 0 contradicting to the
assumption.). Then c/a is rational and nonzero. By ad = bc,
ax + b c
acx + bc
acx + ad
· =
=
= 1,
cx + d a
acx + ad
acx + ad
we have
a
ax + b
= ,
cx + d
c
which is again rational.
To show the other direction, suppose (ax + b)/(cx + d) ∈ Q is given by m/n for some
nonzero integers m, n ∈ Z. Then we have
(am − cn)x = dn − bm,
which means either am − cn = dn − bm = 0 or x = (dn − bm)/(am − cn). If am − cn =
dn − bm = 0, then am = cn and bm = dn. Then we have the following cases
• First, from am = cn, we have a = 0 ⇔ c = 0, in which case ad = bc = 0.
• Similarly, from bm = dn, we have b = 0 ⇔ d = 0, in which case ad = bc = 0.
• At last, none of a, b, c, d is zero. Then a/c = b/d = n/m hence, ad = bc.
(2) (#10) Prove that for all n ∈ N we have
2n − 1
1
1 3 5
· · · ··· ·
≤√
2 4 6
2n
3n + 1
and equality obtains if and only if n = 1.
Proof. We prove by induction. When n = 1, both sides are 1/2 hence the inequality and
the equality.
When the inequality is true for n ∈ N, we want to show
1 3 5
2n + 1
1
· · · ··· ·
≤√
.
2 4 6
2n + 2
3n + 4
Using the inequality for n, this is to prove
1 3 5
2n − 1 2n + 1
1
1
2n + 1
· · · ··· ·
≤√
≤√
.
2 4 6
2n 2n + 2
3n + 1 2n + 2
3n + 4
So we want to show
√
3n + 1
2n + 1
≤√
.
2n + 2
3n + 4
Taking a square of both sides and eventually it is equivalent to show
(3n + 4)(4n2 + 4n + 1) ≤ (3n + 1)(4n2 + 8n + 4)
that is to show
19n ≤ 20n.
MATH 3150
Handout, September 28, 2016
2
Apparently this is true. So the inequality is proved. Also, since 19n < 20n for all n + 1
case where n ∈ N, the equality cannot be obtained for n > 1. Therefore, the equality is
obtained if and only if n = 1.
(3) (#23) If A is an infinite set, then A has a countable infinite subset.
Proof. If A is an infinite set, A is nonempty, choose one element a1 and let B1 = {a1 } ⊂ A.
Assume we have found a subset Bn containing n elements of A for n ∈ N, note that A\Bn
is infinite. (Otherwise, A = Bn ∪ A\Bn is finite, a contradiction.) Hence, we can find an
element in A\Bn , denoted by an+1 , and let Bn+1 = Bn ∪ {an+1 }. Therefore, Bn+1 has n + 1
elements of A. Let
∞
[
B=
Bn .
n=1
We have B = {xn : n ∈ N} is countable. And for any n ∈ N, there are n elements
a1 , a2 , . . . , an in B. Therefore, B is infinite.
(4) (#25)
Proof. a). Since every polynomial of degree n ∈ N has at most n distinct roots in C, to
show that the set of all algebraic numbers is countable, it suffices to show that there are
countably many polynomials with integer coefficients. For each k ∈ N, we consider the
number of polynomials
an z n + an−1 z n−1 + · · · + a1 z + a0
such that
n + |an | + |an−1 | + · · · + |a1 | + |a0 | ≤ k.
Since aj and n are integers less and equal than k, there are at most k + 1 possible values
for n (0, 1, 2, . . . , k) and at most 2k + 1 (including negative integers) possible values of aj .
So there are less than (k + 1)(2k + 1)k+1 (actually much less) polynomials satisfying above
condition. Let
n
n
X
X
Ak = {z ∈ C :
aj z j = 0, n +
|aj | ≤ k}
j=0
j=0
for each k ∈ N. Above shows that Ak is finite hence countable. Then the set of algebraic
∞
[
Ak , a countable union of countable sets, hence is countable.
numbers is
k=0
b). This is true because every rational number m/n (m, n ∈ Z, n 6= 0) is a root of
nz − m = 0, i.e., a1 = n and a0 = −m.
(5) Let A, B ⊂ R, denote
A + B = {a + b : a ∈ A, b ∈ B}.
Show
inf(A + B) = inf A + inf B.
Proof.
• If one of A, B is empty, without loss of generality let A = ∅, then A + B = ∅.
And inf ∅ = ∞. We have
inf(A + B) = ∞ = inf A + inf B.
• Suppose both A and B are nonempty, inf B 6= ∞. If any of A, B is NOT bounded
below, without loss of generality let A unbounded below. Then inf A = −∞. Pick
b ∈ B, for any n ∈ N, there exists an ∈ A such that an ≤ −n − |b|. This means for any
n ∈ N, there exists an element an + b ∈ A + B such that an + b ≤ −n, which proves
inf(A + B) = −∞ = inf A + inf B.
MATH 3150
Handout, September 28, 2016
3
• If both A and B are bounded below, s = inf A and t = inf B are finite. Then a ∈ A,
b ∈ B implies s ≤ a and t ≤ b. Therefore, s + t ≤ a + b. Moreover, for any ε > 0, there
exists a ∈ A and b ∈ B such that a < s + ε/2 and b < t + ε/2. Then a + b < s + t + ε.
This proves
inf(A + B) = s + t = inf A + inf B.
(6) (#18) First, since 1 < 2 < 3 < 5, we have 15 < 31 < 12 < 1. Then
1
1
1
2−(k+1) = 2−k · < 2−k , 3−(k+1) = 3−k · < 3−k , 5−(k+1) = 5−k · < 5−k
2
3
5
for any k ∈ N, and one has
1
1
1
2−k ≤ , 3−k ≤ , 5−k ≤
2
3
5
1
1
1
for any k ∈ N, which shows s = 2 + 3 + 5 is an upper bound of E. Moreover, for any ε > 0,
s ∈ E and s > s − ε. Therefore,
sup E = s.
Next, since 5 > 3 > 2 > 0, by the exponential law (1.24), for any k, m, n ∈ N, we have
3−m and 5−n are positive. Then 0 is a lower bound of E.
Claim: For any k ∈ N, we have 2k ≥ k.
We prove the claim by induction. For k = 1, 2 > 1. Assuming 2k ≥ k, we have
2−k ,
2k+1 = 2k · 2 = 2k + 2k ≥ k + k ≥ k + 1.
This finishes the proof of the Claim.
Similarly, one also has 3m ≥ m and 5n ≥ n for m, n ∈ N.
For any ε > 0, there exists k, m, n ∈ N such that
kε/3 > 1,
mε/3 > 1,
nε/3 > 1.
2−k < ε/3,
3−m < ε/3,
5−n < ε/3.
This further implies
Hence, 2−k + 3−m + 5−n < ε. This proves
inf E = 0.